v_naren said:4) ok...yes you are correct...but there is a way around this problem......what you can do is....place an Ls>0.2nH.........then the real part of input impedance will be wt*Ls>50Ohms...right?!!!...
example say u use an onchip inductor of value= 1.5nH
then with wt=40GHz and Ls=1.5nH
Zin=wt*Ls+ jw(Ls+Lg) + (1/jwCbe)
now u will get wt*Ls=377Ohms....now what u can do is add another matching network between the base of the BJT and the 50Ohm source to convert this 377+j.... to 50Ohm...now u need not use bondwire as u have deliberately used larger Ls and u will be using another simple LC matching network to convert the large 377+j... to 50Ohm. this can be done using the smith chart to design.
I design CMOS LNAs and hence I know the optimum device width and all..but the design procedure is same for both....I dont know if there is an optiumum device size for minimum noise figure for SiGe based LNAs....u cn try Prof.thomas Lee's book for reading CMOS based LNAs using the inductive source degeneration technique....byebye
v_naren said:yes correct...this is the way!!!!!
here is what u need to do....
just make induc src degen stages with as many BJTs as u can....u cannot sweep anything here on!!!!!...
u need to put in BJT1 and Le1 and BJT2 and find Le2....all the time such that Re(Z11)=50Ohm....and always such that the DC power dissipation is a constant with each BJT means u need to change the bias voltage everytime...example..if u say u are going to burn 20mW...then u make sure that the DC power is 20mW with every BJT.
now u need to find the noise figure each time .....the noise figure will reach a minimum for a particular size of BJT. Thats all
now just use this "optimum size" BJT to make ur LNA. with a large inductor at emitter Le...example say you put Le=2nH.
then lets say u get input impedance at base as 200 + j 67 Ohm
now just use a passive lossless L matching network to convert this to 50 +j 0 or rather as 50Ohms.....thats all is the procedure...u can do the band pass thing at the collector of the cascode...and u get a LNA
the theory from the paper is that
for a given DC power dissipation there is only one MOSFET with a certain device width in one technology at any given frequency.
so for a 0.18um CMOS process at 2.45GHz...the MOSFET with Width of 420um will give best least possible noise figure...this is the concept....I just gave u a trick or "technique" to find that optimum size but in a BJT using the simulator itself...ita my own trick...I dont know for a BJT if there exists an optimum device size at all???...cuz I design CMOS LNAs in my research only
v_naren said:420um is for the 0.18um CMOS process which I use...yes correct..it does consume a lot of power and hence u need to set vgs very close to Vth and hence ur input signal swing is limited at the bottom end by vgs-vt.
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